Diffraction and Interference Combined. Consider the...

Diffraction and Interference Combined. Consider the interference pattern produced by two parallel slits of width $a$ and separation $d,$ in which $d=3 a .$ The slits are illuminated by normally incident light of wavelength $\lambda$ . (a) First we ignore diffraction effects due to the slit width. At what angles $\theta$ from the central-maximum will the next four maxima in the two-slit interference pattern occur? Your answer will be in terms of $d$ and $\lambda .$ (b) Now we include the effects of diffraction. If the intensity at $\theta=0$ is $I_{0}$ , what is the intensity at each of the angles in part (a)? (c) Which double-slit interference maxima are missing in the pattern? (d) Compare your results to those illustrated in Fig. 36.12 $\mathrm{c}$ . In what ways is your result different?

Key Concepts

Interference-Diffraction Combination

When both interference and diffraction act simultaneously, the observed intensity pattern is a product of the two effects. The sharp interference fringes from the double-slit setup are modulated by the wider diffraction envelope from each individual slit, leading to variations in fringe intensity and the potential suppression of some interference maxima.

Missing Orders

In the combined pattern, some expected interference maxima may be absent if they occur at angles where the single-slit diffraction envelope has a minimum. This phenomenon, known as missing orders, results from the cancellation effects inherent in the diffraction component completely nullifying the interference signal at those particular angles.

Double-Slit Interference

This concept refers to the formation of fringes resulting from the superposition of light waves emerging from two narrow, closely spaced slits. The condition for constructive interference, which results in bright fringes, is determined by the path difference between the light from the two slits being an integer multiple of the wavelength.

Single-Slit Diffraction

Single-slit diffraction occurs when light passes through a narrow aperture and spreads out, forming a broad diffraction pattern. The pattern is characterized by a central maximum and successive minima and secondary maxima, with the slit width playing a key role in determining the angular positions of these features.

Transcript

00:01 We are told that d equals 3a.

00:08 And i'm using the equation for diffraction of sine theta equals lambda over a.

00:27 And i'm using the equation for interference of lambda equals xd over l.

00:48 Or rather, let me use sine theta equals lambda over d.

01:09 And i can put an m here also.

01:17 There we go.

01:19 So in part a, we're looking for the next four maxima in the pattern.

01:33 So m equals 1, m equals 2, m equals 3, m equals 4.

01:44 Theta is the inverse sign of lambda over d.

01:52 And so putting that in a calculator, setting my calculator for degrees, inverse sign of lambda over d.

02:13 Wait a minute.

02:18 What angles? and yeah, my answer is going to have to be in terms of lambda and d.

02:29 So i can't really put this in a calculator.

02:32 Later, it's just going to have to be theta is the inverse sign of lambda over d, inverse sign of 2 lambda over d, inverse sign 3 lambda over d, inverse sign 4 lambda over d.

03:09 Okay, so that's the answer to a.

03:11 Let's move on to b.

03:19 So we're told to use i sub -zero here.

03:25 What are the intensity of each of those four angles? well, i equals i -subs -0 the sign of 2 pi over the wavelength times a sine theta over 2 pi over wavelength a sine theta squared.

04:16 So the sign of the inverse sign is going to cancel out.

04:47 So we're going to have m lambda over d squared.

05:09 Okay, notice that the wavelengths cancel out.

05:16 So that's going to give me 2 pi a m over d, the sign of 2 pi am over d, squared.

05:41 All right.

06:00 Three, and four.

06:03 C.

06:06 Which double slit interference maximum are missing? okay, so looking at this equation, sine of theta equals lambda over a, and just so that i can differentiate it, i'm going to change this to phi.

06:46 So if the sign of phi equals equals the sign of theta, then a maxima from here will be on a minima from here...

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